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Question Number 68207 by mr W last updated on 07/Sep/19

solve for x∈C sin x=z (z=a+bi=re^(iθ) )

$${solve}\:{for}\:{x}\in\mathbb{C} \\ $$$$\mathrm{sin}\:{x}={z}\:\:\:\left({z}={a}+{bi}={re}^{{i}\theta} \right) \\ $$

Commented by mathmax by abdo last updated on 07/Sep/19

sinx =z ⇔((e^(ix) −e^(−ix) )/(2i))=z ⇔e^(ix) −e^(−ix) =2ir e^(iθ) let t =e^(ix) ⇒ix =ln(t)+i2kπ with k∈Z (e) ⇒t−t^(−1 ) =2ir e^(iθ) ⇒t^2 −1 =2ir e^(iθ) t ⇒ t^2 −2ire^(iθ) t −1 =0 →Δ^′ =(ire^(iθ) )^2 +1 =−r^2 e^(2iθ) +1 t_1 =ir e^(iθ) +(√(1−r^2 e^(2iθ) )) and t_2 =ir e^(iθ) −(√(1−r^2 e^(2iθ) )) ix =ln(t_1 )+i2kπ ⇒−x =ilnt_1 −2kπ ⇒x =−ilnt_1 +2kπ =−iln(r e^(iθ) −i(√(1−r^2 e^(2iθ) )))+2kπ ix =lnt_2 +i2kπ ⇒−x =ilnt_2 −2kπ ⇒x =−ilnt_2 +2kπ =−iln(ir e^(iθ) −(√(1−r^2 e^(2iθ) )))+2kπ

$${sinx}\:={z}\:\Leftrightarrow\frac{{e}^{{ix}} −{e}^{−{ix}} }{\mathrm{2}{i}}={z}\:\Leftrightarrow{e}^{{ix}} −{e}^{−{ix}} =\mathrm{2}{ir}\:{e}^{{i}\theta} \\ $$$${let}\:{t}\:={e}^{{ix}} \:\Rightarrow{ix}\:={ln}\left({t}\right)+{i}\mathrm{2}{k}\pi\:\:\:{with}\:{k}\in{Z} \\ $$$$\left({e}\right)\:\Rightarrow{t}−{t}^{−\mathrm{1}\:} =\mathrm{2}{ir}\:{e}^{{i}\theta} \:\Rightarrow{t}^{\mathrm{2}} −\mathrm{1}\:=\mathrm{2}{ir}\:{e}^{{i}\theta} \:{t}\:\Rightarrow \\ $$$${t}^{\mathrm{2}} −\mathrm{2}{ire}^{{i}\theta} {t}\:−\mathrm{1}\:=\mathrm{0}\:\rightarrow\Delta^{'} =\left({ire}^{{i}\theta} \right)^{\mathrm{2}} \:+\mathrm{1}\:=−{r}^{\mathrm{2}} {e}^{\mathrm{2}{i}\theta} \:+\mathrm{1} \\ $$$${t}_{\mathrm{1}} ={ir}\:{e}^{{i}\theta} \:+\sqrt{\mathrm{1}−{r}^{\mathrm{2}} {e}^{\mathrm{2}{i}\theta} \:}\:\:\:\:{and}\:{t}_{\mathrm{2}} ={ir}\:{e}^{{i}\theta} −\sqrt{\mathrm{1}−{r}^{\mathrm{2}} {e}^{\mathrm{2}{i}\theta} } \\ $$$${ix}\:={ln}\left({t}_{\mathrm{1}} \right)+{i}\mathrm{2}{k}\pi\:\Rightarrow−{x}\:={ilnt}_{\mathrm{1}} −\mathrm{2}{k}\pi\:\Rightarrow{x}\:=−{ilnt}_{\mathrm{1}} +\mathrm{2}{k}\pi \\ $$$$=−{iln}\left({r}\:{e}^{{i}\theta} −{i}\sqrt{\mathrm{1}−{r}^{\mathrm{2}} {e}^{\mathrm{2}{i}\theta} }\right)+\mathrm{2}{k}\pi \\ $$$${ix}\:={lnt}_{\mathrm{2}} +{i}\mathrm{2}{k}\pi\:\Rightarrow−{x}\:={ilnt}_{\mathrm{2}} −\mathrm{2}{k}\pi\:\Rightarrow{x}\:=−{ilnt}_{\mathrm{2}} +\mathrm{2}{k}\pi \\ $$$$=−{iln}\left({ir}\:{e}^{{i}\theta} −\sqrt{\mathrm{1}−{r}^{\mathrm{2}} {e}^{\mathrm{2}{i}\theta} }\right)+\mathrm{2}{k}\pi \\ $$

Answered by Smail last updated on 07/Sep/19

((e^(ix) −e^(−ix) )/2)e^(−iπ/2) =re^(iθ) e^(2ix) −1−2re^(ix) ×e^(i(θ+π/2)) =0 (e^(ix) −re^(i(θ+π/2)) )^2 =1+r^2 e^(2i(θ+π/2)) e^(ix) =+_− (1+r^2 e^(2i(θ+π/2)) )^(1/2) +re^(i(θ+π/2)) x=−iln(re^(i(θ+π/2)) +_− (√(1+r^2 e^(2i(θ+π/2)) ))) x=−isinh^(−1) (re^(i(θ+π/2)) )

$$\frac{{e}^{{ix}} −{e}^{−{ix}} }{\mathrm{2}}{e}^{−{i}\pi/\mathrm{2}} ={re}^{{i}\theta} \\ $$$${e}^{\mathrm{2}{ix}} −\mathrm{1}−\mathrm{2}{re}^{{ix}} ×{e}^{{i}\left(\theta+\pi/\mathrm{2}\right)} =\mathrm{0} \\ $$$$\left({e}^{{ix}} −{re}^{{i}\left(\theta+\pi/\mathrm{2}\right)} \right)^{\mathrm{2}} =\mathrm{1}+{r}^{\mathrm{2}} {e}^{\mathrm{2}{i}\left(\theta+\pi/\mathrm{2}\right)} \\ $$$${e}^{{ix}} =\underset{−} {+}\left(\mathrm{1}+{r}^{\mathrm{2}} {e}^{\mathrm{2}{i}\left(\theta+\pi/\mathrm{2}\right)} \right)^{\mathrm{1}/\mathrm{2}} +{re}^{{i}\left(\theta+\pi/\mathrm{2}\right)} \\ $$$${x}=−{iln}\left({re}^{{i}\left(\theta+\pi/\mathrm{2}\right)} \underset{−} {+}\sqrt{\mathrm{1}+{r}^{\mathrm{2}} {e}^{\mathrm{2}{i}\left(\theta+\pi/\mathrm{2}\right)} }\right) \\ $$$${x}=−{isinh}^{−\mathrm{1}} \left({re}^{{i}\left(\theta+\pi/\mathrm{2}\right)} \right) \\ $$

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